For the chemical equation given below, what is the amount of methane required to produce 72 g of water?
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$

32 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

44 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

64 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
32 g

Mass of water ( $H_{2} O$ ) to be produced = 72 g.
Gram molecular mass of a compound is the sum of gram atomic mass of all the constituent atoms.
Gram molecular mass of water = 18 g
No. of moles of $H_{2} O$ in 72 g
= $\frac{Mass of water}{Gram molecular mass of water}$ = $\frac{72}{18} = 4$ mole

For 2 moles of $H_{2} O$ $\to$ 1 mole of methane is required
So, for 4 moles of $H_{2} O$ $\to$ 2 moles of methane is required.
Gram molecular mass of methane = 16 g
The required amount of methane $= 2 \times 16 = 32 g$ .

Suggest Corrections

Similar questions

C H_{4} (_{g}) + 2 O_{2} (_{g}) \to C O_{2} (_{g}) + 2 H_{2} O (_{g})

H_{2} O_{(g)}

8.00 g

C H_{4}

C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}

H_{2}

C H_{4} (g) + \frac{1}{2} O_{2} (g) \to C O (g) + 2 H_{2} (g)

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g); △ H = - 800 k J

C H_{4} (g) + C O_{2} (g) \to 2 C O (g) + 2 H_{2} (g); △ H = + 235 k J

C H_{4} (g) + H_{2} O (g) \to C O (g) + 3 H_{2} (g); △ H = + 204 k J

△ H

What change is represented by the following equation?

CH₄(g) + 20₂(g) →CO₂(g) + 2H₂O(l)

C (s) + O_{2} (g) \to C O_{2} (g) + 94.2 K c a l

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) + 68.3 K c a l

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) + 210.8 K c a l

Join BYJU'S Learning Program